Comparing pressure of air in a tire installed in a car & a free tire

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The pressure of air in a tire installed on a car (P_c) is slightly higher than in a free tire (P_f) due to the weight of the car compressing the tire, which reduces its volume and increases pressure. However, this difference is minimal and typically undetectable on a tire gauge. For example, with a car weighing 4000 pounds and each tire supporting 1000 pounds, the contact patch only slightly alters the tire's volume and pressure. Therefore, when inflating a tire to 30 psi, it is acceptable to inflate it to that same pressure even when it is not mounted on the axle. The pressure increase upon installation is negligible, confirming that the same pressure can be used for both scenarios.
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Imagine an ordinary car wheel with rubber tire, installed in a car axle. So the weight of the car (some of the weight of course, because there are 4 wheels) is pressing on it. Let's denote this pressure of the air in the tire as P_{c}.

Now the same wheel (with its rim and tire) remove from the car axle, so it is now freely lying on the floor. Hence no car weight is pressing on it. Let's denote the pressure of the air in the tire as P_{f}.

Question: Is P_{c} different than P_{f} ? How to show it by using ideal gas law equation? You may assume the weight of the rim as negligible if needed.

My personal guess: P_{c} is more than P_{f} because with the weight of the car pressing on the tire, volume of tire (hence volume of gas) reduces, thus pressure increases. In other words, if a tire pressure is 30psi while installed on car, once take it out from the car axle, its pressure will drop.
 
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You are correct, but the difference is so small that you will not notice it on a tire gauge.

Suppose your car is 4000 pounds with 1000 pounds on each wheel. Also suppose your tires are 10 inches wide. At 30 psi your contact patch for each wheel will be about 33 square inches. Since the tires are 10 inches wide that means your contact patch is only 3.3 inches long. Deforming the tire with a typical 6.5+ foot circumfrence so that there is a 3.3 inch flat spot will change it's volume (and pressure) only very slightly.
 
mrspeedybob said:
You are correct, but the difference is so small that you will not notice it on a tire gauge.

Suppose your car is 4000 pounds with 1000 pounds on each wheel. Also suppose your tires are 10 inches wide. At 30 psi your contact patch for each wheel will be about 33 square inches. Since the tires are 10 inches wide that means your contact patch is only 3.3 inches long. Deforming the tire with a typical 6.5+ foot circumfrence so that there is a 3.3 inch flat spot will change it's volume (and pressure) only very slightly.

Hence, if I was to pump air to a tire but I can only carry a free wheel instead of driving the car to the air pump, says I wish to have 30psi in my tire when it is installed on axle, then I should just directly pump in 30psi instead of a slightly lower pressure (for instance, 29psi) because the increase in pressure after the tire is installed is just so insignificant?
 
Correct.
 

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